# Talk:Cavalieri's principle

## Spheres and Cone Example

This example states that

the plane located y units above the "equator" intersects the sphere in a circle of area ${\displaystyle \pi (r^{2}-y^{2})}$

And that

the area of the plane's intersection with the part of the cylinder that is outside of the cone is also ${\displaystyle \pi (r^{2}-y^{2})}$

Can someone double check that? I am trying to follow through the example and it seems that the area of the intersection circle in the sphere should be ${\displaystyle 2\pi {\sqrt {r^{2}-y^{2}}}}$ and the intersection ring of the inner cylinder outside of the ring should be ${\displaystyle 2\pi (r-y)}$, different from what is claimed. —Preceding unsigned comment added by 192.223.226.6 (talkcontribs)

The article's statements are correct and yours are not. The circumference of the intersection of the plane with the sphere is 2π√(r2 − y2), i.e. it's 2π times the radius. The area is π times the square of the radius. Since the radius is √(r2 − y2), the area is π(r2 − y2).
Remember that the area needs to be proportional to the squares of the distances involved, and that fails to happen in either of the two formulas you give.
Similarly in the cone example, what you give is the difference between the circumferences rather than the difference between the areas of the two concentric circles. Michael Hardy (talk) 20:22, 28 May 2009 (UTC)
Ah, that was very silly of me. Thanks for the clairification and the quick response. 192.223.226.6 (talk) 13:20, 29 May 2009 (UTC)

### Missing working, missing triangle

I had to read the same proof several times until I realized that a whole lot of working is missing in the jump to Pythagoras, which is asserted but not demonstrated. It would help if it was shown on the diagram that it is the annulus we are interested in (it should be shaded) and similarly that the slice through the sphere (shaded the same way) has that same area -- we need the triangle from the centre of the sphere to the centre of the slice to a point on the slice's circumference. Then Pythagoras appears, and then even I can follow the rest of the proof. Chiswick Chap (talk) 08:44, 24 October 2013 (UTC)

## Possible change

I think we should change this:

"It is a short step from there to the conclusion that the area under a single whole cycloidal arch is three times the area of the circle. Which then means that the area of a rectangle bounding one half of a single cycloidal arch is two times the area of the circle, the area of a rectangle bounding a single whole cycloidal arch is four times the area of the circle, and the rectangularly-bounded area above a single whole cycloidal arch is exactly equal to the area of the circle.[citation needed]"

to this:

"It is a short step from there to the conclusion that the area under a single whole cycloidal arch is three times the area of the circle. (In addition, if you draw a rectangle around a cycloidal arch, the area of the portion of the rectangle _above_ the arch is exactly equal to that of the circle.)"

Also, why the "citation needed"? You can follow the logic by yourself.

## Date of Formulation?

Shouldn't the article make some mention of the (perhaps approximate) date on which the principle was formulated (in Europe)? Did Cavalieri come up with it, or is it just now named after him? The article states: "Cavalieri's principle was originally called the method of indivisibles, the name it was known by in Renaissance Europe." Was it known in Renaissance Europe before him (and perhaps he merely formalized it, clarified it, or otherwise popularized it), or was it known there because of him? If the former, then when was it first formulated in Europe? If the latter, when did he come up with it? Either way it should be mentioned, even if we can give a no better answer than "in the 17th century" (the time during which Cavalieri lived). Does anyone know the details? — Preceding unsigned comment added by Shawn81 (talkcontribs) 15:19, 31 August 2015 (UTC)